Solution Set of System of ODE. I am trying to find the solution of the system
$$\begin{bmatrix}x_1\\x_2\end{bmatrix}'=  \begin{bmatrix}1&3\\3&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}$$.
I am given that \begin{bmatrix} e^{4t} \\ e^{4t} \end{bmatrix} and \begin{bmatrix} e^{-2t} \\ -e^{-2t} \end{bmatrix} are solutions, and any linear combination is as well: $$\begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}\Bigg(A\begin{bmatrix} e^{4t} \\ e^{4t} \end{bmatrix}+B\begin{bmatrix} e^{-2t} \\ -e^{-2t} \end{bmatrix}\Bigg)=\begin{bmatrix} 4Ae^{4t}-2Be^{-2t} \\ 4Ae^{4t}+2Be^{-2t} \end{bmatrix}=\begin{bmatrix} Ae^{4t}+Be{-2t} \\ Ae^{4t}-Be^{-2t} \end{bmatrix}'$$ So the solution set is infinite.
On the other hand, from linear algebra, shouldn't the solution set be just 1 element since the homogenous equation $\begin{bmatrix}1&3\\3&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ has only the trivial solution $x_1=x_2=0$?
 A: I don't think there is relevance between the number of solutions of the linear system $Ax = b$ and that of the system of ODE $x' = Ax$.
For the system of ODE $x' = Ax$, there must be $n$ linearly independent solutions arouse from the eigenvalues of $A$, irrespective of the rank of $A$.
For completeness, I would like to add that the only relation between the numbers of solutions of $Ax = b$ and $x' = Ax$ I can think of is:
$$
\begin{split}
& Ax = b \text{ has a unique solution } \\
\iff &  \det(A) \neq 0\\
\iff & 0 \text{ is not an eigenvalue of } A\\
\iff & \text{for any solution of } x' = Ax, y = P^{-1}x \text{ could not be a constant function}\\
\end{split}
$$ 
where $A = PJP^{-1}$ is the Jordan decomposition. Nevertheless, there are still infinite solutions for $x' = Ax$, no matter how many solutions $Ax = b$ has.
A: \begin{equation*}
\mathbf{x}^{\prime }=\mathbf{W\cdot x\Rightarrow x}(t)=\exp [\mathbf{W}%
t]\cdot \mathbf{x}(0),\;\mathbf{W}=\left(
\begin{array}{cc}
1 & 3 \\
3 & 1
\end{array}
\right)
\end{equation*}
We can diagonalise $\mathbf{W}$. Thus
\begin{equation*}
\mathbf{W}=\left(
\begin{array}{cc}
1 & 3 \\
3 & 1
\end{array}
\right) =\mathbf{U}^{-1}\mathbf{\cdot }\left(
\begin{array}{cc}
4 & 0 \\
0 & -2
\end{array}
\right) \cdot \mathbf{U}=\mathbf{U}^{-1}\cdot \mathbf{D\cdot U}
\end{equation*}
where $\mathbf{U}$ is a unitary matrix (which can be calculated if so
desired).
\begin{eqnarray*}
\mathbf{x}(t) &=&\exp [\mathbf{W}t]\mathbf{x}(0)=\mathbf{U}^{-1}\mathbf{
\cdot }\exp [\mathbf{D}t]\cdot \mathbf{U\cdot x}(0) \\
\mathbf{U\cdot x}(t) &=&\exp [\mathbf{D}t]\cdot \mathbf{x}(0),\;\exp [
\mathbf{D}t]=\left(
\begin{array}{cc}
e^{4t} & 0 \\
0 & e^{-2t}
\end{array}
\right)  \\
\lbrack \mathbf{U\cdot x}(t)]_{1} &=&e^{4t}[\mathbf{U\cdot x}(0)]_{1},\;[
\mathbf{U\cdot x}(t)]_{2}=e^{4t}[\mathbf{U\cdot x}(0)]_{2}
\end{eqnarray*}
This is the most general solution.
